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QUESTION


\begin{description}

\item[(a)]
Fine the vector equation and standard equation of the following
planes:

\begin{description}

\item[(i)]
The plane $\Pi_1$ which passes through the point $P=(1,2,3)$ and
has the normal vector
$\mathbf{n}=2\mathbf{i}-1\mathbf{j}+3\mathbf{k}$;

\item[(ii)]
The plane $\Pi_2$ which passes through the three points
$P_1=(8,8,-1),\ P_2=(2,0,-8),\ P_3=(14,30,4)$;

\item[(iii)]
The plane $\Pi_3$ which passes through the point $Q=(5,0,-8)$ and
is parallel to the plane with equation $6x-3y+9z=11$.

\item[(iv)]
Are any of the planes $\Pi_1,\ \Pi_2,\ \Pi_3$ parallel? Are any of
them equal?

\end{description}

\item[(b)]
Let $\mathbf{a}=2\mathbf{i}+2\mathbf{j}-\mathbf{k}$ and
$\mathbf{b}=2\mathbf{i}-\mathbf{j}$.

\begin{description}

\item[(i)]
Find the relation that must hold between $x_1,x_2$ and $x_3$ if
the vector $\mathbf{x}=x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}$
is to be written as $\mathbf{x}=s\mathbf{a}+t\mathbf{b}$ where $s$
and $t$ are scalars.

\item[(ii)]
Show that the vector
$\mathbf{c}=2\mathbf{i}+8\mathbf{j}-3\mathbf{k}$ can be written as
$\mathbf{c}=s\mathbf{a}+t\mathbf{b}$.

\item[(iii)]
Fins $s$ and $t$ in this case.

\end{description}

\end{description}



ANSWER


\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
$$\mathbf{v}. \left(\begin{array}{c}2\\-1\\3\end{array}\right)
=\left(\begin{array}{c}2\\-1\\3\end{array}\right).
\left(\begin{array}{c}1\\2\\3\end{array}\right)=9$$

$$2x-y+3z=9$$

\item[(ii)]
$\mathbf{u}=\vec{P_2P_1}=6\mathbf{i}+8\mathbf{j}+7\mathbf{k}$

$\mathbf{v}=\vec{p_2P_3}=12\mathbf{i}+30\mathbf{j}+12\mathbf{k}$

\begin{eqnarray*}
\mathbf{n}=\mathbf{u}\times\mathbf{v}&amp;=&amp;
\left|\begin{array}{ccc}\mathbf{i}&amp;\mathbf{j}&amp;\mathbf{k}
\\6&amp;8&amp;7\\12&amp;30&amp;12\end{array}\right|\\
&amp;=&amp;\left|\begin{array}{cc}8&amp;7\\30&amp;12\end{array}\right|
\mathbf{i}-\left|\begin{array}{cc}6&amp;7\\12&amp;12\end{array}
\right|\mathbf{j}+\left|\begin{array}{cc}6&amp;8\\12&amp;30\end{array}\right|\mathbf{k}\\
&amp;=&amp;(96-210)\mathbf{i}-(72-84)\mathbf{j}+(-180-96)\mathbf{k}\\
&amp;=&amp;-114\mathbf{i}+12\mathbf{j}+84\mathbf{k}
\end{eqnarray*}

This can be scaled to $-19\mathbf{i}+12\mathbf{j}+84\mathbf{k}$

\begin{eqnarray*}
\mathbf{w.n}=\vec{OP_2}.\mathbf{n}&amp;=&amp;\left(\begin{array}{c}2\\0\\-8
\end{array}\right).\left(\begin{array}{c}-114\\12\\84\end{array}\right)\\
&amp;=&amp;-228-672=900\\ -114x+12y+84z&amp;=&amp;900
\end{eqnarray*}

\item[(iii)]
$\mathbf{u}.\mathbf{n}=\left(\begin{array}{c}5\\0\\-8
\end{array}\right).\mathbf{n}$ where
$\mathbf{n}=6\mathbf{i}-3\mathbf{j}+9\mathbf{k}$

$$6x-3y+9z=-42$$

\item[(iv)]
To check for parallelism check normals: First and third are
parallel. Since equations are inconsistent they are distinct.

\end{description}

\item[(b)]

\begin{description}

\item[(i)]
$\mathbf{x}.(\mathbf{a}\times\mathbf{b})=0$

$$\mathbf{a}\times\mathbf{b}=\left|\begin{array}{ccc}
\mathbf{i}&amp;\mathbf{j}&amp;\mathbf{k}\\2&amp;2&amp;-1\\2&amp;-1&amp;0
\end{array}\right|=-\mathbf{i}-2\mathbf{j}-6\mathbf{k}$$

so $x_1+2x_2+6x_3=0$

\item[(ii)]
$\mathbf{c}.(\mathbf{a}\times\mathbf{b})=-2-16+2+18=0$

\item[(iii)]
$\mathbf{c}\times\mathbf{a}=-t(\mathbf{a}\times\mathbf{b})$ and
$\mathbf{c}\times\mathbf{b}=s(\mathbf{a}\times\mathbf{b})$

$$\mathbf{c}\times\mathbf{a}=\left|\begin{array}{ccc}
\mathbf{i}&amp;\mathbf{j}&amp;\mathbf{j}\\2&amp;8&amp;-3\\2&amp;2&amp;-1\end{array}
\right|=-2\mathbf{i}-4\mathbf{j}-12\mathbf{k}$$

so $t=-2$

$$\mathbf{c}\times\mathbf{b}=\left|\begin{array}{ccc}
\mathbf{i}&amp;\mathbf{j}&amp;\mathbf{k}\\2&amp;8&amp;-3\\2&amp;-1&amp;0\end{array}
\right|=-3\mathbf{i}-6\mathbf{j}-18\mathbf{k}$$

so $s=3$

\end{description}

\end{description}




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